Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $z \neq 0$. $t = \dfrac{-9z^2 - 63z}{z - 10} \div \dfrac{z^2 + 10z + 21}{7z - 70} $
Dividing by an expression is the same as multiplying by its inverse. $t = \dfrac{-9z^2 - 63z}{z - 10} \times \dfrac{7z - 70}{z^2 + 10z + 21} $ First factor the quadratic. $t = \dfrac{-9z^2 - 63z}{z - 10} \times \dfrac{7z - 70}{(z + 7)(z + 3)} $ Then factor out any other terms. $t = \dfrac{-9z(z + 7)}{z - 10} \times \dfrac{7(z - 10)}{(z + 7)(z + 3)} $ Then multiply the two numerators and multiply the two denominators. $t = \dfrac{ -9z(z + 7) \times 7(z - 10) } { (z - 10) \times (z + 7)(z + 3) } $ $t = \dfrac{ -63z(z + 7)(z - 10)}{ (z - 10)(z + 7)(z + 3)} $ Notice that $(z - 10)$ and $(z + 7)$ appear in both the numerator and denominator so we can cancel them. $t = \dfrac{ -63z\cancel{(z + 7)}(z - 10)}{ (z - 10)\cancel{(z + 7)}(z + 3)} $ We are dividing by $z + 7$ , so $z + 7 \neq 0$ Therefore, $z \neq -7$ $t = \dfrac{ -63z\cancel{(z + 7)}\cancel{(z - 10)}}{ \cancel{(z - 10)}\cancel{(z + 7)}(z + 3)} $ We are dividing by $z - 10$ , so $z - 10 \neq 0$ Therefore, $z \neq 10$ $t = \dfrac{-63z}{z + 3} ; \space z \neq -7 ; \space z \neq 10 $